find the sum of first 40 integers divisible by 6
Answers
Answered by
6
hey your answer is 4920
=6+12+18+24......240.
6(1+2+3+4+.....40)
{sum of 1to 10=55}
{sum of 11 to 20=155}
{sum of 21to 30=255}
{sum of 31to40=355}
so answer is 6(55+155+255+355)
6(820)=4920
=6+12+18+24......240.
6(1+2+3+4+.....40)
{sum of 1to 10=55}
{sum of 11 to 20=155}
{sum of 21to 30=255}
{sum of 31to40=355}
so answer is 6(55+155+255+355)
6(820)=4920
saadu91:
tq
Answered by
7
6,12,18,24,30,36,42,48...240
This forms an Arithmetic Progression
a=First Term = 6
Common Difference=d= 6
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